Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:

import java.util.*;

public class Main
{
    private final static int MOD = 1_000_000_007;
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
	int N = Integer.parseInt(in.nextLine());
	double[] arr = new double[N];
	String[] temp = in.nextLine().split(" ");
	for (int i = 0; i < N; i++) {
	    arr[i] = Double.parseDouble(temp[i]);
	}
	double[][] dp = new double[N+1][N+1];
        dp[0][0] = 1;
	for (int i = 1; i <= N; i++) {
	    dp[i][0] = dp[i-1][0] * (1 - arr[i-1]);
	    for (int j = 1; j <= i; j++) {
		dp[i][j] += dp[i-1][j-1] * arr[i-1];
		dp[i][j] += dp[i-1][j] * (1 - arr[i-1]);
	    }
	}
	double res = 0;
	for (int i = 0; i <= N; i++) {
	    if (i > N-i) {
		res += dp[N][i];
	    }
	}
	System.out.println(res);
	in.close();
    }
}